Difference between revisions of "2023 AMC 12A Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
− | Think about <math>2^3+4^3+6^3+...+18^3</math>. Once we factor out <math>2^3=8</math>, we get <math>1^3+2^3+...+9^3</math>, | + | Think about <math>2^3+4^3+6^3+...+18^3</math>. Once we factor out <math>2^3=8</math>, we get <math>1^3+2^3+...+9^3</math>, which can be found using the sum of cubes formula, <math>(\frac{n(n+1)}{2})^2</math>. Now think about <math>1^3+3^3+...+17^3</math>. This is just the previous sum subtracted from the total sum of <math>18</math> cubes. So now we have the two things we need to add. The sum of all the even cubes is <math>8\cdot (\frac{90}{2})^2\rightarrow 8\cdot 2025 = 16200</math>. The sum of all cubes from <math>1^3</math> to <math>18^3</math> is <math>(\frac{18\cdot 19}{2})^2=29241</math>. The sum of the odd cubes is then <math>29241-16200=13041</math>. Thus we get <math>16200-13041=\boxed{\textbf{(D) } 3159}</math> |
+ | ~amcrunner | ||
+ | |||
+ | ==Solution 2 (a bit faster)== | ||
+ | Using the same sum of cubes formula, we can rewrite as <math>2(2^3 + 4^3 + ... + 18^3) - (1^3 + 2^3 + ... + 18^3)</math> | ||
+ | |||
+ | <math>= 2(2^3)(1^3 + ... + 9^3) - (1^3 + ... + 18^3)</math> | ||
+ | |||
+ | <math>= 16(5 \cdot 9)^2 - (9 \cdot 19)^2 = 9^2(20^2 - 19^2) = 81 \cdot 39 = \boxed{\textbf{(D) } 3159}</math> | ||
+ | ~AoPSuser216 | ||
==Solution 3== | ==Solution 3== | ||
− | For any real numbers x and y, <math>x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)((x-y)^2+3xy)=(x-y)^3+ | + | For any real numbers <math>x</math> and <math>y</math>, <math>x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)((x-y)^2+3xy)=(x-y)^3+3xy(x-y)</math>. |
+ | |||
+ | When <math>x = y + 1</math>, with the above formula, we will get <math>x^3-y^3=1^3+3xy=1 + 3xy</math>. | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <math>2^3 - 1^3 + 4^3 - 3^3 + \dots + 18^3 - 17^3</math> | ||
+ | |||
+ | <math>= (1 + 3\cdot 1\cdot 2) + (1 + 3\cdot 3\cdot 4) + \dots + (1 + 3\cdot 17\cdot 18)</math> | ||
+ | |||
+ | <math>= 9 + 3\cdot (1\cdot 2 + 3\cdot 4 + \dots + 17\cdot 18)</math> | ||
+ | |||
+ | <math>= 9 + 3 \cdot (2 + 12 + 30 + 56 + 90 + 132 + 182 + 240 + 306)</math> | ||
+ | |||
+ | <math>= 9 + 3 \cdot 1050</math> | ||
− | + | <math>= \boxed{\textbf{(D) } 3159}</math> | |
~sqroot | ~sqroot | ||
+ | |||
+ | Alternatively, to avoid the long sum, | ||
+ | |||
+ | <math> (1\cdot 2 + 3\cdot 4 + \dots + 17\cdot 18) \\ | ||
+ | = (2^2)(9(10)(19)/6) - (2)(9(10)/2) \\ | ||
+ | =(2)(9)(10)(2(19/6) - 1/2) \\ | ||
+ | = 180(35/6) = 35(30) = 1050</math> | ||
+ | |||
+ | ==Solution 4== | ||
+ | We rewrite the sum as | ||
+ | |||
+ | <cmath>\sum_{k=1}^{9}(2k)^3-(2k-1)^3</cmath> | ||
+ | <cmath>=\sum_{k=1}^{9} 12k^2 - 6k + 1</cmath> | ||
+ | <cmath>=12\sum_{k=1}^{9}k^2 - 6\sum_{k=1}^{9}k + 9</cmath> | ||
+ | <cmath>=12\cdot\frac{9\cdot 10\cdot 19}{6} -6\cdot \frac{9\cdot 10}{2} + 9</cmath> | ||
+ | <cmath>=3420 - 270 + 9 = \boxed{\textbf{(D) } 3159}</cmath> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | ==Solution 4 (answer choices)== | ||
+ | We see <math>=12\cdot\frac{9\cdot 10\cdot 19}{6} -6\cdot \frac{9\cdot 10}{2} + 9 = 9\cdot (12\cdot\frac{10\cdot 19}{6} -6\cdot \frac{10}{2} + 1)</math> which is clearly a multiple of 9. The only answer choice which is a multiple of 9 is <math>\boxed{\textbf{(D) } 3159}</math> ~Ilaggo2432 | ||
+ | |||
+ | ==Solution 5 (Bash)== | ||
+ | |||
+ | <math> 2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3</math> | ||
+ | |||
+ | <math>=8-1+64-27+216-125+512-343+1000-729+1728-1331+2744-2197+4096-3375+5832-4913</math> | ||
+ | |||
+ | <math>= \boxed{\textbf{(D) } 3159}</math> | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Reduce all terms mod 9. This yields: | ||
+ | |||
+ | <math> 2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3</math> | ||
+ | <math> \equiv -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 +0 - -1 (\mod 9)</math> | ||
+ | <math> \equiv 0 (\mod 9)</math> | ||
+ | |||
+ | The only answer choice which is also ≡0 mod 9 is <math>= \boxed{\textbf{(D) } 3159}</math> | ||
+ | |||
+ | ==Video Solution == | ||
+ | |||
+ | by Power Solve | ||
+ | |||
+ | https://youtu.be/YXIH3UbLqK8?si=RZhSDIKjRNLrgVS5&t=1552 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/33Tz-bfKzmw | ||
+ | <i> ~Education, the Study of Everything </i> | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/_eoPL5H8b0k | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2023|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2023|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 07:26, 24 August 2024
Contents
Problem
What is the value of
Solution 1
To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas.
we could rewrite the second part as
Hence,
Adding everything up:
~lptoggled
Solution 2
Think about . Once we factor out
, we get
, which can be found using the sum of cubes formula,
. Now think about
. This is just the previous sum subtracted from the total sum of
cubes. So now we have the two things we need to add. The sum of all the even cubes is
. The sum of all cubes from
to
is
. The sum of the odd cubes is then
. Thus we get
~amcrunner
Solution 2 (a bit faster)
Using the same sum of cubes formula, we can rewrite as
~AoPSuser216
Solution 3
For any real numbers and
,
.
When , with the above formula, we will get
.
Therefore,
~sqroot
Alternatively, to avoid the long sum,
Solution 4
We rewrite the sum as
-Benedict T (countmath1)
Solution 4 (answer choices)
We see which is clearly a multiple of 9. The only answer choice which is a multiple of 9 is
~Ilaggo2432
Solution 5 (Bash)
Solution 6
Reduce all terms mod 9. This yields:
The only answer choice which is also ≡0 mod 9 is
Video Solution
by Power Solve
https://youtu.be/YXIH3UbLqK8?si=RZhSDIKjRNLrgVS5&t=1552
Video Solution
https://youtu.be/33Tz-bfKzmw ~Education, the Study of Everything
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.